Question

the freezing point of the solution containing 0.3 g of acetic acid in 30 g of benzene is lowered by 0.45 C . Calculate Vant Hoff factor. Kf for benzene (c6h6)=5.12kg /mol

Answer 1

Answer: 0.52

Explanation:

Formula used for lowering in freezing point is,

$\Delta T_f=i\times k_f\times m$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

$\Delta T_f=i\times k_f\times \frac{\text{mass in g}}{\text{ Molecular mass}\times \text{Mass of solvent in kg}}$

$0.45=i\times 5.12kg/mol\times \frac{0.3g}{\text{ Molecular mass}\times {0.03kg}}$

$0.45=i\times 5.12kg/mol\times \frac{0.3g}{60g/mol}\times {0.03kg}$

$i=0.52$